If the work done in blowing a soap bubble of volume ' $\mathrm{V}$ ' is ' $\mathrm{W}$ ', then the work done…
If the work done in blowing a soap bubble of volume ' $\mathrm{V}$ ' is ' $\mathrm{W}$ ', then the work done in blowing a soap bubble of volume ' $2 \mathrm{~V}$ ' will be
$2 \mathrm{~W}$
$4^{1 / 3} \mathrm{~W}$
W
$\sqrt{2} \mathrm{~W}$
Solution
If $\mathrm{V}$ is the volume then we have
$\begin{aligned}
& \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \\
& \mathrm{~V} \propto \mathrm{r}^3 \\
& \therefore \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^3 \\
& \frac{\mathrm{r}_2}{\mathrm{r}_1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{1 / 3}=(2)^{1 / 3} \\
& \mathrm{~W}_1=8 \pi \mathrm{r}_1^2 \cdot \mathrm{T} \text { and } \mathrm{W}_2=8 \pi \mathrm{r}_2^2 \mathrm{~T} \\
& \therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=(2)^{2 / 3}=(4)^{1 / 3} \\
& \mathrm{~W}_2=(4)^{1 / 3} \mathrm{~W}
\end{aligned}$